Question

find the exact value of sec(pi/8)

Answer 1

\[\frac{1}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8}\]

Answer 2

1/2*pi/4 means
half of pi/4
think half angle identity

Answer 3

I'm confused where the half angle identity comes in

Answer 4

well we know pi/4 is on the unit circle
and half of pi/4 is pi/8
so
since pi/8 is in the first quadrant we know cos to be positive
\[\cos(\frac{\pi}{8}) \\ =\cos(\frac{1}{2} \cdot \frac{\pi}{4})=\sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}}\]

Answer 5

I haven't seen that equation before

Answer 6

\[\cos(\frac{\theta}{2})= \pm \sqrt{\frac{1+\cos(\theta)}{2}}\]?
you haven't seen this one?

Answer 7

have you seen this one:
\[\cos^2(\theta)=\frac{1}{2}(1+\cos(2 \theta))\]

Answer 8

maybe I wrote it down for the wrong thing

Answer 9

My notes are all wrong lol

Answer 10

http://www.purplemath.com/modules/idents.htm
here are a lot of trig identities with the corresponding names

Answer 11

maybe you can use this to fix your notes

Answer 12

\[\cos \frac{ x }{ 2 }=\sqrt{\frac{ 1+sinx }{ 2 }}\] that's what I have

Answer 13

+- in the front

Answer 14

err yes that sin(x) thing is definitely suppose to be cos(x)

Answer 15

Apex is terrible

Answer 16

\[\cos(\frac{x}{2})= \pm \sqrt{\frac{1+\cos(x)}{2}} \\ x=\frac{\pi}{4} \\ \cos(\frac{\pi/4}{2})=\pm \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \\ \cos(\frac{\pi}{8}) \text{ is positive since } \frac{\pi}{8} \text{ is \in the first quadrant } \\ \cos(\frac{\pi}{8})= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \sec(\frac{\pi}{8})=\frac{1}{\cos(\frac{\pi}{8})} =\sqrt{\frac{2}{1+\cos(\frac{\pi}{4})}} \]

Answer 17

there is still a couple things for you to do

Answer 18

find cos(pi/4) using unit circle

Answer 19

and simplify

Answer 20

how'd you get cos pi/4 ?

Answer 21

x is pi/4

Answer 22

i replace x with pi/4

Answer 23

\[\frac{1}{2} \cdot \frac{\pi}{4} =\frac{\pi}{8} \\ \text{ so } x=\frac{\pi}{4}\]

Answer 24

oooh okay

Answer 25

cos pi/4 is sqrt2/2

Answer 26

and 45 degrees

Answer 27

right so you can replace cos(pi/4) is sqrt(2)/2

Answer 28

well pi/4 is 45 deg

Answer 29

\[\cos(\frac{x}{2})= \pm \sqrt{\frac{1+\cos(x)}{2}} \\ x=\frac{\pi}{4} \\ \cos(\frac{\pi/4}{2})=\pm \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \\ \cos(\frac{\pi}{8}) \text{ is positive since } \frac{\pi}{8} \text{ is \in the first quadrant } \\ \cos(\frac{\pi}{8})= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \sec(\frac{\pi}{8})=\frac{1}{\cos(\frac{\pi}{8})} =\sqrt{\frac{2}{1+\cos(\frac{\pi}{4})}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}}\]

Answer 30

yeah I didn't know if I needed the degrees or radians so I put both lol

Answer 31

you don't need either you just need to find cos(45) or cos(pi/4)

Answer 32

cos(45 deg) is equivalent to finding cos(pi/4)

Answer 33

do I need to get rid of the radical or leave that there ?

Answer 34

I would get rid of the compound fraction action inside the radical and then see if there are any perfect squares afterwards

Answer 35

so would I multiply top and bottom by the opposite of the denominator ?

Answer 36

\[\sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}} \cdot \sqrt{\frac{2}{2}} \\ \sec(\frac{\pi}{8}) = \sqrt{ \frac{2 (2)}{(1+\frac{\sqrt{2}}{2})(2)}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{4}{2+\sqrt{2}}}\]
notice the numerator inside the squareroot is a perfect square

Answer 37

\[\sec(\frac{\pi}{8})=\frac{\sqrt{4}}{\sqrt{2+\sqrt{2}}}\]

Answer 38

wouldn't the rad 2 or rad 2 equal 1 ?

Answer 39

yes sqrt(2/2)=1
that is why I multiplied it to get rid of the compound fraction action inside the square root

Answer 40

I'm confused on the multiplying of 2 in the denominator and numerator

Answer 41

let's look inside the square root
we have inside the square root:
\[\frac{2}{1+\frac{\sqrt{2}}{2}}\]
this has a mini-fraction the sqrt(2)/2
the sqrt(2)/2 has a 2 in the denominator
I multiply top and bottom inside the square root by 2

so I have inside the square root:
\[\frac{2}{1+\frac{\sqrt{2}}{2}} \cdot \frac{2}{2} \\ \frac{2(2)}{(1+\frac{\sqrt{2}}{2})(2)} \\ \frac{4}{1(2)+\frac{\sqrt{2}}{2}(2)} \\ \frac{4}{2+ \frac{2}{2} \sqrt{2}} \\ \\ \frac{4}{2+(1) \sqrt{2}} \\ \frac{4}{2+ \sqrt{2}}\]

Answer 42

why did you transfer the radical to the two ?

Answer 43

what 2 ?

Answer 44

are you asking me about this:
\[\frac{\sqrt{2}}{2}(2)=\frac{2}{2} \sqrt{2}\]

Answer 45

this part

Answer 46

multiplication is commuative

Answer 47

\[\frac{\sqrt{2}}{2}(2) =\frac{\sqrt{2}}{2} \cdot \frac{2}{1} = \frac{\sqrt{2} \cdot 2 }{2 \cdot 1 } \\ \text{ you change order \in multiplication } \\ =\frac{2 \sqrt{2}}{2 \cdot 1} \\ \text{ you can reseperate the fraction } \\ =\frac{2}{2} \frac{\sqrt{2}}{1} \\ =1 \frac{\sqrt{2}}{1} \\ 1 \cdot \sqrt{2} \ = \sqrt{2}\]

Answer 48

oooh okay , that makes sense

Answer 49

all I did was use the fact that a*b is the same as b*a

Answer 50

okay, that makes a lot of sense

Answer 51

for my final answer , I got \[-\sqrt{2}+2\] is that right ?

Answer 52

\[\cos(\frac{x}{2})= \pm \sqrt{\frac{1+\cos(x)}{2}} \\ x=\frac{\pi}{4} \\ \cos(\frac{\pi/4}{2})=\pm \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \\ \cos(\frac{\pi}{8}) \text{ is positive since } \frac{\pi}{8} \text{ is \in the first quadrant } \\ \cos(\frac{\pi}{8})= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \sec(\frac{\pi}{8})=\frac{1}{\cos(\frac{\pi}{8})} =\sqrt{\frac{2}{1+\cos(\frac{\pi}{4})}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}} \\ \]
\[\sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}} \cdot \sqrt{\frac{2}{2}} \\ \sec(\frac{\pi}{8}) = \sqrt{ \frac{2 (2)}{(1+\frac{\sqrt{2}}{2})(2)}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{4}{2+\sqrt{2}}}\]
\[\sec(\frac{\pi}{8})=\frac{\sqrt{4}}{\sqrt{2+\sqrt{2}}}\]
\[\sec(\frac{\pi}{8})=\frac{2}{\sqrt{2+\sqrt{2}}}\]

Answer 53

wouldn't the rad 2 cancel out since it's two radicals ?

Answer 54

so it'd be \[\frac{ 2 }{ \sqrt{2}+2 }\]

Answer 55

no \[\sqrt{a+\sqrt{b}} \neq \sqrt{a}+b \\ \text{ take } a=4 \text{ and } b=16 \\ \text{ so } \sqrt{a}=2 \text{ and } \sqrt{b}=4 \\ \sqrt{a+\sqrt{b}}=\sqrt{4+4}=\sqrt{8} \\ \text{ while } \sqrt{a}+b=2+16=18 \\ \text{ and } 18 \text{ is definitely not equal to } \sqrt{8} \\ \text{ so } \sqrt{a+\sqrt{b}} \text{ is not } \sqrt{a}+b\]

Answer 56

so it stays like that ?

Answer 57

it stays like \[\frac{2}{\sqrt{2+\sqrt{2}}}\]
yes
you could rationalize the bottom but it will still look ugly

Answer 58

so I would leave it as is

Answer 59

I never learned that lol . I have to get ready for work now though , thank you for the help !

Answer 60

good luck

Answer 61

If I post up a new question with this activity I have to do , do you think you could figure it out ? I tried but I couldn't get it . I got stuck

Answer 62

The activities that are worth points in this packet are super confusing